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- Feb 14, 2012

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Solve for $x$ in terms of $a$ in the equation $x^4-3ax^3+3a^3x+a^4=0$.

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- Feb 14, 2012

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Solve for $x$ in terms of $a$ in the equation $x^4-3ax^3+3a^3x+a^4=0$.

- Aug 30, 2012

- 1,178

Great. Now I'm not getting any sleep tonight.

-Dan

That wasn't as bad as I thought. A little guess work and serendipity go a long way.

Let's start with the (almost preposterous) idea that it can be factored:

\(\displaystyle x^4 - 3ax^3 + 3a^3x + a^4 = (x^2 + bx + c)(x^2 + dx + e)\)

Expanding out gives use the equations:

\(\displaystyle b + d = -3a\)

\(\displaystyle c^2 + bd + e^2 = 0\)

\(\displaystyle be + cd = 3a^3\)

\(\displaystyle ce = a^4\)

Let's try \(\displaystyle c = e = a^2\).

Then we get, from the first and third equations that \(\displaystyle b + d = -3a\) and \(\displaystyle b + d = 3a\). No good.

Let's try again. Let \(\displaystyle c = e = -a^2\).

Then the first and third equations both read \(\displaystyle b + d = -3a\). So we now have \(\displaystyle d = -3a - b\).This means our last equation (the second one) now gives

\(\displaystyle b^2 + 3ab + 2a^2 = (b + 2a)(b + a) = 0\)

So we have that b = -2a or b = -a.

Both of them give the same answer, so let's put b = -2a.

Tallying up:

\(\displaystyle b = -2a\)

\(\displaystyle c = -a^2\)

\(\displaystyle d = -a\)

\(\displaystyle e = -a^2\).

Thus (whew!)

\(\displaystyle x^4 - 3ax^3 + 3a^3x + a^4 = (x^2 - 2ax - a^2)(x^2 - ax - a^2) = 0\)

This solves down to the following four solutions:

\(\displaystyle x = a ( 1 \pm \sqrt{2} )\)

\(\displaystyle x = \frac{a}{2} ( 1 \pm \sqrt{5} )\)

-Dan

-Dan

Let's start with the (almost preposterous) idea that it can be factored:

\(\displaystyle x^4 - 3ax^3 + 3a^3x + a^4 = (x^2 + bx + c)(x^2 + dx + e)\)

Expanding out gives use the equations:

\(\displaystyle b + d = -3a\)

\(\displaystyle c^2 + bd + e^2 = 0\)

\(\displaystyle be + cd = 3a^3\)

\(\displaystyle ce = a^4\)

Let's try \(\displaystyle c = e = a^2\).

Then we get, from the first and third equations that \(\displaystyle b + d = -3a\) and \(\displaystyle b + d = 3a\). No good.

Let's try again. Let \(\displaystyle c = e = -a^2\).

Then the first and third equations both read \(\displaystyle b + d = -3a\). So we now have \(\displaystyle d = -3a - b\).This means our last equation (the second one) now gives

\(\displaystyle b^2 + 3ab + 2a^2 = (b + 2a)(b + a) = 0\)

So we have that b = -2a or b = -a.

Both of them give the same answer, so let's put b = -2a.

Tallying up:

\(\displaystyle b = -2a\)

\(\displaystyle c = -a^2\)

\(\displaystyle d = -a\)

\(\displaystyle e = -a^2\).

Thus (whew!)

\(\displaystyle x^4 - 3ax^3 + 3a^3x + a^4 = (x^2 - 2ax - a^2)(x^2 - ax - a^2) = 0\)

This solves down to the following four solutions:

\(\displaystyle x = a ( 1 \pm \sqrt{2} )\)

\(\displaystyle x = \frac{a}{2} ( 1 \pm \sqrt{5} )\)

-Dan

Last edited:

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- Feb 14, 2012

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HiGreat. Now I'm not getting any sleep tonight.

-Dan

I take this as you like this challenge problem and I wish you the best of luck to crack it it no time!

- Aug 30, 2012

- 1,178

Actually, I did. A little motivation helps...I have to be up by 6.HiDan,

I take this as you like this challenge problem and I wish you the best of luck to crack it it no time!

-Dan

- Mar 31, 2013

- 1,346

my solution

X^4 – 3ax^3 + 3a^3 x + a^4 = 0

Forgetting the sign it is symmetric from left to right and a is increasing as x is decresing

So divide by x^2 and collect x^n and 1/x^n together

(X^2 + a^4/x^2) – 3a(x - a^2 /x) = 0

Or (x-a^2/x)^2 + 2a^2 – 3a(x-a^2/x) = 0

Putting x- a^2/x = y we get y^2 + 2a^2 – 3ay = 0

So y = a or 2a

So x^2 - 2ax – a^2 = 0 or x^2 –ax – a^2 = 0

both quadratic can be solved

Forgetting the sign it is symmetric from left to right and a is increasing as x is decresing

So divide by x^2 and collect x^n and 1/x^n together

(X^2 + a^4/x^2) – 3a(x - a^2 /x) = 0

Or (x-a^2/x)^2 + 2a^2 – 3a(x-a^2/x) = 0

Putting x- a^2/x = y we get y^2 + 2a^2 – 3ay = 0

So y = a or 2a

So x^2 - 2ax – a^2 = 0 or x^2 –ax – a^2 = 0

both quadratic can be solved

Last edited:

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- Feb 14, 2012

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Thanks for participating,my solution

Forgetting the sign it is symmetric from left to right and a is increasing as x is decresing

So divide by x^2 and collect x^n and 1/x^n together

(X^2 + a^4/x^2) – 3a(x - a^2 /x) = 0

Or (x-a^2/x)^2 + 2a^2 – 3a(x-a^2/x) = 0

Putting x- a^2/x = y we get y^2 + 2a^2 – 3ay = 0

So y = a or 2a

So x^2 - 2ax – a^2 = 0 or x^2 –ax – a^2 = 0

both quadratic can be solved

But I don't quite agree with the end result that you obtained.

Take for example if \(\displaystyle x^2 - 2ax – a^2 = 0\) is true, solving it for $x$ we get:

\(\displaystyle x=a\pm a\sqrt{2}=a(1\pm \sqrt{2})\)

Substitute the expression of $x=a(1+\sqrt{2})$ above into the original given equation, we have:

\(\displaystyle x^4 – 3ax^3 + 3a^3 x + a^4=(a(1+ \sqrt{2}))^4 – 3a(a(1+ \sqrt{2}))^3 + 3a^3(a(1+ \sqrt{2})+ a^4=-12a^4 \ne 0\)

What do you think,

- Mar 31, 2013

- 1,346

I did not check you calculation.Thanks for participating,kaliprasad...

But I don't quite agree with the end result that you obtained.

Take for example if \(\displaystyle x^2 - 2ax – a^2 = 0\) is true, solving it for $x$ we get:

\(\displaystyle x=a\pm a\sqrt{2}=a(1\pm \sqrt{2})\)

Substitute the expression of $x=a(1+\sqrt{2})$ above into the original given equation, we have:

\(\displaystyle x^4 – 3ax^3 + 3a^3 x + a^4=(a(1+ \sqrt{2}))^4 – 3a(a(1+ \sqrt{2}))^3 + 3a^3(a(1+ \sqrt{2})+ a^4=-12a^4 \ne 0\)

What do you think,kaliprasad?

my solution matches with that of topsquark and further

(x^2-2ax - a^2)(x^2-ax - a^2) gives me the product

(x^4 - 3ax^3 + 3a^3 x +a4)

where is the mistake

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- #8

- Feb 14, 2012

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First, I didn't see

And thanks for participating in this problem to both of you!